Question

How do I find the derivative of `f(x) = sqrt(x+3)` using first principles?

Answer 1

`f'(x)=1/(2sqrt(x+3))`

Explanation:

`f'(x)=lim_(h->0)(f(x+h)-f(x))/h`

`f(x)=sqrt(x+3), f(x+h)=sqrt(x+h+3)`, then

`f'(x)=lim_(h->0)(sqrt(x+h+3)-sqrt(x+3))/h`

If we evaluate this right away, we get

`lim_(h->0)(sqrt(x+h+3)-sqrt(x+3))/h=(sqrt(x+3)-sqrt(x+3))/0=0/0`,

so we need to simplify as this is an indeterminate form.

Multiply the entire limit by the numerator's conjugate, which is `(sqrt(x+h+3)+sqrt(x+3))/(sqrt(x+h+3)+sqrt(x+3))`. This is the same as multiplying by `1.`

`f'(x)=lim_(h->0)(sqrt(x+h+3)-sqrt(x+3))/h*(sqrt(x+h+3)+sqrt(x+3))/(sqrt(x+h+3)+sqrt(x+3))`

The numerator becomes

`sqrt(x+h+3)-sqrt(x+3) * [sqrt(x+h+3)+sqrt(x+3)]=x+h+3-(x+3)=x+h+3-x-3=h`

`f'(x)=lim_(h->0)(cancelx+h+cancel3-cancelx-cancel3)/(h(sqrt(x+h+3)+sqrt(x+3))`

`f'(x)=lim_(h->0)(cancelh)/(cancelh(sqrt(x+h+3)+sqrt(x+3))`

`f'(x)=lim_(h->0)1/(sqrt(x+h+3)+sqrt(x+3))`

`f'(x)=1/(sqrt(x+3)+sqrt(x+3))`

`f'(x)=1/(2sqrt(x+3))`